Okay, dq = dA = 2r dr. dA = 2r dr because a ring with inner radiusR 1 and outer

Question

Okay, dq = dA = 2r dr. dA = 2r dr because a ring with inner radiusR1 and outer radius R2 has an area of A= r(2)2 -r(1)2. Using grouping, A =(r2+r1) r , where r =r2-r1 . For small r, r1˜ r2 and r2 + r1 ˜r where r < r < r2. So, for small r, A˜ 2r r. Then dA = 2r dr . From all that, I have to show thatr(2)2 -r(1)2 = ( r2 +r1 ) r. Thanks for any and all help!!! Okay, dq = dA = 2r dr. dA = 2r dr because a ring with inner radiusR1 and outer radius R2 has an area of A= r(2)2 -r(1)2. Using grouping, A =(r2+r1) r , where r =r2-r1 . For small r, r1˜ r2 and r2 + r1 ˜r where r < r < r2. So, for small r, A˜ 2r r. Then dA = 2r dr . From all that, I have to show thatr(2)2 -r(1)2 = ( r2 +r1 ) r. Thanks for any and all help!!!

Explanation / Answer

Given :         dq = dA= 2r dr. dA = 2r dr because a ring with inner radiusR1 and outer radius R2 has an area of A= r(2)2 -r(1)2             Now take A = r22 -r12                                     = ( r22   - r12 )                                    = ( r2 + r1 )(  r2  - r1)    since a2 -b2 = ( a + b) ( a - b)                                   = ( r2 + r1 ) r        since r is nothing but chane in radius  =   r2-r1 Hope this helps u!

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